Existence and attractivity of solutions for fractional difference equations

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R E S E A R C H

Open Access

Existence and attractivity of solutions for

fractional difference equations

Lu Zhang

1

_{and Yong Zhou}

1,2*

*_{Correspondence:}

[email protected] 1_{Faculty of Mathematics and}

Computational Science, Xiangtan University, Xiangtan, P.R. China 2_{Nonlinear Analysis and Applied}

Mathematics (NAAM) Research Group, Faculty of Science, King Abdulaziz University, Jeddah, Saudi Arabia

Abstract

In this paper, we study a kind of difference equations with Riemann–Liouville-like fractional difference. The results on existence and attractivity are obtained by using the Picard iteration method and Schauder’s fixed point theorem. Examples are provided to illustrate the main results.

MSC: 39A06; 39A13

Keywords: Fractional diﬀerence equations; Existence; Attractivity

1 Introduction

In this paper, we study the fractional diﬀerence equation

⎧ ⎨ ⎩

α

α–1u(t) =f(t+α,u(t+α)), t∈N0,α∈(0, 1],

α_α–1_–1u(t)|t=0=u0,

(1.1)

whereα _{is the Riemann–Liouville-like fractional diﬀerence, which is deﬁned in later} sections,f :Nα×R→R,f(t,u) is continuous with respect totandu, andNα={α,α+ 1,α+ 2, . . .}.

In the last decade, fractional differential equations have been recognized as valuable tools to describe many phenomena in various fields of engineering, physics, science, and so on. A huge number of results focused on fractional differential equations; see the monographs of Kilbas et al. [12] and Zhou [17, 18], the papers [1, 10], and the references therein. Within the past ten years, however, there has been more interest in developing discrete fractional equations, that is, fractional difference equations. This development has demonstrated that fractional difference equations have a number of unexpected diffi-culties and technical complications [2–6, 9, 13–15].

Motivated by the works mentioned, in this paper, we investigate the existence and at-tractivity of solutions for fractional difference equations. In Sect. 2, we describe the dis-crete fractional difference calculus and some properties. The main results are obtained in Sect. 3. Using the Picard iteration method, we prove the existence of equation (1.1) in Sect. 3.1. In Sect. 3.2, we also obtain the existence of attractive solutions for fractional difference equations. Section 3.3 is devoted to inducing the existence and attractivity of equation (1.1) by Schauder’s fixed point theorem according to the introduced weighted space. Finally, we provide three examples to illustrate our results.

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2 Preliminaries

This section is devoted to some preliminary facts.

Deﬁnition 2.1(see [5, 11]) Letν> 0. Theνth fractional sum is deﬁned by

–_aνf(t) = 1 (ν)

t–ν

s=a

t–σ(s)(ν–1)f(s), (2.1)

where f is deﬁned fors=a mod(1), and–ν

a f is deﬁned fort=a+ν mod(1), t(ν)= (t+1)

(t–ν+1),σ(s) =s+ 1, andis the gamma function. The fractional sum–

ν_{maps functions} deﬁned onNato functions deﬁned onNa+ν.

Definition 2.2(see [5, 11]) Letμ> 0 andm– 1≤μ≤m, wheremis a positive integer. Theμth fractional difference is defined as

_aμf(t) =m_a–(m–μ)f(t). (2.2)

Lemma 2.1(see [4]) Assume thatμ+ 1is not a nonpositive integer.Then

–_aνt(μ)₌ (μ+ 1)

(μ+ν+ 1)t

(μ+ν)_. _(2.3)

Lemma 2.2(see [4]) Assume that the following factorial functions are well deﬁned.Then: (i) t(μ)₌_μ_t(μ–1)_;

(ii) (t–μ)t(μ)₌_t(μ+1)_;

(iii) μ(μ)₌_(μ_{+ 1);}

(iv) t(μ+ν)_{= (}_t_–_ν)(μ)_t(ν)_.

We indicate the following properties of the gamma function (see [7, 12]): (i) for allx∈R, excludingx= 0, –1, –2, . . .,

(x)(1 –x) = π

sin(πx);

(ii) the gamma function is logarithmically convex function on the positive real axis, that is,log(x)is convex.

By the log-convexity property of the gamma function we have that

λx+ (1 –λ)y≤λ(x)1–λ(y), λ∈(0, 1),x,y> 0.

Lemma 2.3

(i) If0 <α< 1,then(t(ν)₎α_≤_t(αν)_;

(ii) Ifα> 1andν> 0,then

t(–ν)α≤(t– [t] +αν) α₍_t_{– [}_t_{] +}_ν)t

(–αν) _for_t_∈_R+_.

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Proof The proof of (i) is given in [4]. Chen [8] proved (ii) fort∈N1. We prove (ii) for all

t∈R+_:

α–1(t+ 1)(t+αν+ 1)α_t_{– [}_t_{] +}_ν

=tα–1(t– 1)α–1· · ·t– [t]α–1(t+αν)(t– 1 +αν)· · ·t– [t] +αν

×t– [t] +αναt– [t] +ν

=

tα 1 +αν t

(t– 1)α _{1 +} αν t– 1

· · ·t– [t]α 1 + αν (t– [t])

×t– [t] +αναt– [t] +ν

<

tα _{1 +}ν t

α

(t– 1)α _{1 +} ν t– 1

α

· · ·t– [t]α 1 + ν (t– [t])

α

×t– [t] +αναt– [t] +ν

= (t+ν)α(t– 1 +ν)α· · ·t– [t] +νααt– [t] +νt– [t] +αν

=α(t+ν+ 1)t– [t] +αν,

where the inequality 1 +αν t < (1 +

ν t)

α_{is used. Then}

α–1₍_t_{+ 1)}

α₍_t₊_ν_{+ 1)}≤

(t– [t] +αν) α₍_t_{– [}_t_{] +}_ν)·

1 (t+αν+ 1).

Hence

t(–ν)α

=

α₍_t_{+ 1)} α₍_t₊_ν_{+ 1)}≤

(t– [t] +αν) α₍_t_{– [}_t_{] +}_ν)·

(t+ 1) (t+αν+ 1)=

(t– [t] +αν) α₍_t_{– [}_t_{] +}_ν)t

(–αν)

fort∈R+_{. The proof is completed.}

Remark2.1 In (ii), ift– [t] +αν∈(0, 1), then we also have that

t(–ν)α≤t(–αν).

Lemma 2.4(see [12, (1.5.15)]) The quotient expansion of two gamma functions at inﬁnity is

(z+a) (z+b)=z

a–b _{1 +}_O 1

z arg(z+a)<π,|z| → ∞

. (2.4)

By Lemma 2.4 we can easily get that

t(–ν)→0 ast→ ∞(ν> 0) (2.5)

and

t(a)

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Lemma 2.5

(i) If–1 <t<r,thent(–ν)_≥_r(–ν)_for_ν_{> 0;}

(ii) Ifν– 1 <t<r,thent(ν)_≤_r(ν)_for_ν_{> 0;}

(iii) Ifα,β≥0,thent(β)_t(–α)_≤_t(β–α)_for_t_>_β_{– 1.}

Proof (i) Since

t(–ν)

r(–ν) =

(t+ 1)(r+ν+ 1) (t+ν+ 1)(r+ 1)

= (t+ 1)(r+ν+ 1)

(λ(r+ν+ 1) + (1 –λ)(t+ 1))((1 –λ)(r+ν+ 1) +λ(t+ 1)),

whereλ= ν

r–t+ν∈(0, 1), by the log-convexity property of the gamma function, we get that

t(–ν)

r(–ν) =

(t+ 1)(r+ν+ 1)

(λ(r+ν+ 1) + (1 –λ)(t+ 1))((1 –λ)(r+ν+ 1) +λ(t+ 1))

≥ (t+ 1)(r+ν+ 1) ((r+ν+ 1))λ+1–λ₍₍_t_{+ 1))}1–λ+λ

= 1.

Thent(–ν)_≥_r(–ν)_{for –1 <}_t_<_r_and_ν_{> 0.}

(ii) Since

t(ν)

r(ν) =

(t+ 1)(r–ν+ 1) (t–ν+ 1)(r+ 1),

similarly to (i), we can easily get thatt(ν)_≤_r(ν)_for_ν_{– 1 <}_t_<_r_and_ν_{> 0.}

(iii) By (i) we know that

t(–α)_≤₍_t_–_β₎(–α) _for_t_>_β_{– 1.}

According to (iv) of Lemma 2.2, it is easy to get that

t(β)t(–α)≤t(β)(t–β)(–α)=t(β–α) fort>β– 1.

The proof is completed.

Remark2.2 Chen and Liu [9] obtained (i) of Lemma 2.5. However, the proof holds in the case thatt→ ∞. Then we give the proof as before.

Lemma 2.6(see [3]) Let a,ν∈R,a>ν> –1,and a≤b.Then b

s=a

s(ν)=(b+ 1)

(ν+1)_–_a(ν+1)

ν+ 1 . (2.7)

In particular,if a=ν,then

b

s=ν

s(ν)=(b+ 1)

(ν+1)

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Remark2.3 Letν> 0 be noninteger, and letm– 1≤μ≤m, wheremis a positive integer. Setν=m–μ. Then we have

t–μ

s=a+ν

t–σ(s)(μ–1)=(t–a–ν)

(μ)

μ . (2.9)

Lemma 2.7(see [4]) Let f be a real-value function deﬁned onNa,and letμ,ν> 0.Then we have the following equalities:

(i) –ν a+μ(

–μ

a f(t)) =–(aμ+ν)f(t) =–aμ+ν(–aνf(t)); (ii) –ν

a f(t) =–aνf(t) –

(t–a)(ν–1) (ν) f(a).

Definition 2.3 The discrete Mittag-Leffler function is defined by

Fα,β(λ,t) = ∞

n=0

λn t

(nα)

(nα+β)

|λ|< 1,

whereα,β∈R+andλ∈C.

Notice that the series is absolutely convergent for|λ|< 1. In fact, since

₍_nt_α(nα₊)_β)= (t+ 1) (t–nα+ 1)(nα+β)

=(t+ 1)(nα–t)sin(π(t–nα+ 1)) π (nα+β)

≤ 1 π

(t+ 1)(nα–t) (nα+β)

≤ 1 π

(t+ 1)((t+ 1))η₍₍_n_α₊_β))1–η (nα+β)

= 1 π

((t+ 1))1+η ((nα+β))η

≤((t+ 1))1+η π

forη= t+1

nα+β–t–1 andn> 2t+1

α , we have

∞

n>2t_α+1 λn t

(α)

(nα+β)

≤((t+ 1))1+η π

∞

n>2t_α+1

|λ|n.

It is easy to see that the series∞_n₌₀λn t(α)

(nα+β) (|λ|< 1) is absolutely convergent.

Deﬁnition 2.4 A solutionuof the fractional diﬀerence equation (1.1) is said to be attrac-tive ifu(t)→0 ast→ ∞.

The space∞_n₀ is the set of real sequences deﬁned on the set of positive integers where any individual sequence is bounded with respect to the usual supremum norm. It is well known that, under the supremum norm,∞_n

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Deﬁnition 2.5 (see [16]) A set of sequences in ∞_n

0 is uniformly Cauchy (or equi-Cauchy) if for every ε> 0, there exists an integer Nsuch that|x(i) –x(j)|<εwhenever i,j>Nfor anyx={x(n)}in.

Theorem 2.1(see [16, Discrete Arzelà–Ascoli’s theorem]) A bounded,uniformly Cauchy

subsetof∞_n₀is relatively compact.

Deﬁnition 2.6 Let (X,d) be a metric space. An operatorT:X→Xis a Picard operator

if there existsx∗∈Xsuch thatFixT={x∗}and the sequence{Tn₍_x

0)}n∈Nconverges tox∗

for allx0∈X.

Theorem 2.2(Schauder’s ﬁxed point theorem) Letbe a closed,convex,and nonempty

subset of a Banach space X.Let T:→be a continuous operator such that Tis a relatively compact subset of X.Then T has at least one ﬁxed point in.

According to Deﬁnitions 2.1–2.2, it is suitable to rewrite the fractional diﬀerence equa-tions (1.1) in the equivalent summation equation

u(t) = u0 (α)t

(α–1)₊ 1

(α) t–α

s=0

t–σ(s)(α–1)fs+α,u(s+α), t∈Nα. (2.10)

3 Main results

3.1 Results via Picard iteration method

Theorem 3.1 Assume that

(H1) f(t,u)satisﬁes the Lipschitz condition

f(t,u1) –f(t,u2)≤A|u1–u2|, (3.1)

where0 <A< 1is independent oft;

(H2) there exists a constantM> 0such that|f(t,u(t))| ≤Mfort∈Nα.

Then the fractional diﬀerence equation(1.1)has at least one solution.

Proof Deﬁne the sequence{gn(·) :n∈N0}as follows:

g0(t) =

u0

(α)t

(α–1)_, _t_∈_N_α, _(3.2)

gn(t) =g0(t) +

1 (α)

t–α

s=0

t–σ(s)(α–1)fs+α,gn–1(s+α)

, t∈Nα,n∈N0. (3.3)

Clearly, by induction we have

gn(t) –gn–1(t)≤MAn–1

t(nα)

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In fact, forn= 1, by condition (H1) we can conclude that

g1(t) –g0(t)≤

1 (α)

t–α

s=0

t–σ(s)(α–1)M

=M t

(α)

(α+ 1).

Without loss of generality, we assume that

gn–1(t) –gn–2(t)≤MAn–2

t((n–1)α)

((n– 1)α+ 1).

Then

_gn(t) –gn–1(t)≤A

1 (α)

t–α

s=0

t–σ(s)(α–1)MAn–2 (s+α)

((n–1)α)

((n– 1)α+ 1)

= MA

n–1

((n– 1)α+ 1)

–α

(t+α)((n–1)α)

=MAn–1 t

(nα)

(nα+ 1).

Set

g(t) = lim

n→∞gn(t) =nlim→∞

gn(t) –g0(t)

+g0(t) = ∞

k=1

gk(t) –gk–1(t)

+g0(t).

Since the series M_A∞_k₌₁Ak₍t_k(k_αα₊₁₎) =M_AFα,β(A,t) is absolutely convergent for 0 <A< 1, the

existence of the solution for the fractional diﬀerence equation (1.1) is proved. The proof

is completed.

3.2 Results via Schauder’s ﬁxed point theorem

In this section, we deal with the existence and attractivity of the solution for fractional difference equations by a fixed point theorem. First, for anyu∈∞_α , we define the operator T as follows:

Tu(t) = u0 (α)t

(α–1)₊ 1

(α) t–α

s=0

t–σ(s)(α–1)fs+α,u(s+α), t∈Nα. (3.4)

Then the existence of the solution of the fractional diﬀerence equation (1.1) is equivalent to thatThas a ﬁxed point.

Theorem 3.2 Assume that

(H3) there exist constantsβ1∈(α, 1),δ> 0,andL1≥0such that

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Furthermore,suppose that

|u0|

(α)α

(α–1)₊ L1(1 –β1)

(1 +α–β1)

α(α–β1)≤_1.

Then the fractional diﬀerence equation(1.1)has at least one solution in S1.Further,the

solutions of (1.1)are attractive.

Proof Chooseγ> 0 suﬃciently small such that

α+γ– 1 < 0, 1 –β1–γ δ> 0, α+γ–β1–γ δ< 0,

and

|u0|

(α)(α+γ)

(α+γ–1)₊ L1(1 –β1–γ δ)

(1 +α–β1–γ δ)

(α+γ)(α–β1+γ–γ δ)≤_1.

Fort∈Nα, deﬁne the closed subsetS1⊂∞α as follows:

S1=

u∈∞_α :u(t)≤t(–γ)fort∈Nα

.

It is easy to see thatS1is a closed, bounded, and convex subset of the Banach space∞α. In the caseδ > 1, we can get the following inequality by (ii) of Lemma 2.3 and Re-mark 2.1:

t(–γ)δ≤(α+γ δ) δ_(α₊_γ₎t

(–γ δ)_≤_t(–γ δ) _for_t_∈_N_α.

Combining this with (i) of Lemma 2.3, we have

t(–γ)δ≤t(–γ δ) forδ> 0,t∈Nα. (3.6)

We ﬁrst show thatTis continuous inS1and mapsS1intoS1. Applying (H3), (3.4), and

(3.6) to anyu∈S1, we have

Tu(t)≤ |u0| (α)t

(α–1)₊ 1

(α) t–α

s=0

t–σ(s)(α–1)fs+α,u(s+α)

≤ |u0|

(α)t

(α–1)₊ 1

(α) t–α

s=0

t–σ(s)(α–1)L1(s+α+ 1)(–β1)u(s+α)

δ

≤ |u0|

(α)t

(α–1)₊ 1

(α) t–α

s=0

t–σ(s)(α–1)L1(s+α+γ δ)(–β1)u(s+α)

δ

≤ |u0|

(α)t

(α–1)₊ L1

(α) t–α

s=0

t–σ(s)(α–1)(s+α+γ δ)(–β1)₍_s₊_α)(–γ)δ

≤ |u0|

(α)t

(α–1)₊ L1

(α) t–α

s=0

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≤ 2L1

Hence we can get that

Tun–Tu =sup

Therefore{Tu:u∈S1}is a bounded and uniformly Cauchy subset by Deﬁnition 2.5. Hence

TS1is relatively compact by Theorem 2.1. Due to Schauder’s ﬁxed point theorem,Thas

a ﬁxed point. All functions inS1tend to 0 ast→ ∞, and hence the solution of (1.1) is

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3.3 Results in a weighted space

Let∞_α be the set of all real sequencesu={u(t)}∞_t₌α with the norm u =supt∈Nα|u(t)|. Then (∞α, · ) is a Banach space. For anyu∈∞α, we introduce a new norm as follows:

u β2=sup

t∈Nα _|

u(t)| t(β2)

,

whereβ2∈(α, 1). It is easy to see that (∞α , · β2) is also a Banach space. Deﬁne the closed subset

Sβ2=

u∈∞_α : u β2≤

(α|u0|+M)(α+ 1 –β2)

(α+ 1)

.

Obviously,Sβ2is a bounded, closed, and convex subset of the Banach space (∞α, · β2).

Deﬁnition 3.1 A functionf : [0,∞)×R→Risβ2-continuous if for anyε> 0, there exists

δ=δε> 0 such that for all (t,u1), (t,u2)∈[0,∞)×R, we have f(t,u1) –f(t,u2)<ε,

provided that |u1–u2| t(β2) <δ.

Obviously,uis a solution of (1.1) if and only if the operatorThas a ﬁxed point.

Theorem 3.3 Assume that(H2)holds and f isβ2-continuous. Then the fractional

dif-ference equation(1.1)has at least one solution in Sβ2.Further,the solutions of (1.1)are attractive.

Proof We ﬁrst show thatTis continuous inSβ2and mapsSβ2intoSβ2. Applying (3.4), for anyu∈Sβ2, we have

|Tu(t)| t(β2) ≤

|u0|t(α–1)

(α)t(β2) +

1 (α)t(β2)

t–α

s=0

t–σ(s)(α–1)fs+α,u(s+α)

≤ |u0|t(α–1)

(α)t(β2) + M (α)t(β2)

t–α

s=0

t–σ(s)(α–1)

= |u0|t

(α–1)

(α)t(β2) +

Mt(α)

(α+ 1)t(β2)

= |u0|(t–β2+ 1) (α)(t–α+ 2)+

M(t–β2+ 1)

(α+ 1)(t–α+ 1)

≤ (α|u0|+M)(α+ 1 –β2)

(α+ 1) ,

which implies thatTmapsSβ2intoSβ2.

Letun,u∈Sβ2,n= 1, 2, . . . , andlimn→∞un=u, which means that un–u β2 →0 as n→ ∞, that is,

sup

t∈Nα

|un(t) –u(t)|

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Combining this with (2.9), we have

Hence we get that

Tun–Tu β2=sup

In the following, we prove thatTSβ2 is relatively compact. To this end, we deﬁne

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+ M

Therefore,{Hu:u∈Sβ2} is a bounded and uniformly Cauchy subset by Deﬁnition 2.5. Thus{Tu(t)

t(β2) :u∈Sβ2}is relatively compact by Theorem 2.1. HenceT(Sβ2) is relatively com-pact inSβ2. Due to Schauder’s ﬁxed point theorem,Thas a ﬁxed point. All functions inSβ2 tend to 0 ast→ ∞, and hence the solution of (1.1) is attractive. The proof is completed.

4 Example

As applications of our main results, we consider the following examples.

Example4.1 Consider the equation

⎧

(H1) and (H2) hold. By Theorem 3.1 we get that (4.1) has at least one attractive solution.

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wheref(t,u(t)) = (t+ 0.2)(–0.7)_u1.2₍_t_),_t_∈_N

0.5. Sinceα= 0.5,β1= 0.7,δ= 1.2,γ = 0.2, we

have that

α+γ– 1 < 0, 1 –β1–γ δ> 0, α+γ–β1–γ δ< 0, (4.6)

and

f(t,u)≤L1(t+γ δ)(–β1)|u|δ fort∈N0.5andu∈∞0.5. (4.7)

Then (H3) holds. By Theorem 3.2 we get that (4.5) has at least one attractive solution.

Example4.3 Consider

⎧ ⎨ ⎩

0.5_–0.5u(t) = (t+ 0.7)(–0.7)sin((t+ 0.5)(–0.6)u(t+ 0.5)), t∈N0,

–0.5

–0.5u(t)|t=0= 0,

(4.8)

wheref(t,u(t)) = (t+ 0.2)(–0.7)_sin₍_t(–0.6)_u₍_t_)),_t_∈_N

0.5. Letβ2= 0.6∈(0.5, 1). Since

f(t,u1) –f(t,u2)=(t+ 0.2)(–0.7)

sint(–0.6)u1

–sint(–0.6)u2

≤(1.7) (2.4)t

(–0.6)_t(0.6)|u1–u2|

t(0.6)

≤(1.7) (2.4)

|u1–u2|

t(0.6) (4.9)

and

f(t,u)=(t+ 0.2)(–0.7)sint(–0.6)u(t)≤(1.7)

(2.4) fort∈N0.5andu∈

∞

0.5. (4.10)

Thenf isβ2-continuous, and (H2) holds. By Theorem 3.3 we get that (4.8) has at least one

attractive solution.

Acknowledgements

The authors are very grateful to the reviewers for their valuable comments. The work was supported by the National Natural Science Foundation of China (No. 11671339).

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

Both authors LZ and YZ contributed equally to each part of this work. Both authors read and approved the ﬁnal manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations.

Received: 31 January 2018 Accepted: 9 May 2018

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